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Cloud [144]
3 years ago
11

The energy of a given wave in the electromagnetic spectrum is 2.64 × 10-21 joules, and the value of Planck’s constant is 6.6 × 1

0-34 joule·seconds. What is the value of the frequency of the wave?
Physics
2 answers:
Lyrx [107]3 years ago
8 0

Answer:

4.00 × 1012 hertz

Explanation:

olga_2 [115]3 years ago
4 0
20x=10 - ion kno i think
You might be interested in
Calculate the potential at the center of curvature of the arc if the potential is assumed to be zero at infinity.
yan [13]

the potential at the center of curvature of the arc = v = Q ∕ (4πε∘a) or \frac{kQ}{a}

ATQ,

We have density of charge,

λ = \frac{Q}{L}

Where L is the rod's length, in this case the semicircle's length L = πr

Q is the charge on the rod

The potential created at the center by an differential element of charge is:

k = \frac{dQ}{r}

where k is the coulomb's constant

r is the distance from dQ to center of the circle

v = ∫ \frac{K dQ}{a} , Where a = radius, k = 1 / 4πε∘

v =\frac{kQ}{a} or Q ∕ (4πε∘a)

To learn more about potential from the given link

brainly.com/question/25923373

#SPJ4

6 0
2 years ago
Convert 10 kilometers to feet.
ehidna [41]

Answer:

32,808.4 feet.

Explanation:

1 kilometer is equal to 3,280.84 feet.

3,280.84 feet × 10 feet = 32,808.4 feet

8 0
3 years ago
I need help on this question please answer within like a PLEASE QUICK HELP
Ludmilka [50]
The answer is c 5 starrrrrr
7 0
3 years ago
A dune buggy moves through the hallway at 61.5 cm/s. How far does it travel in 4 minutes?
Contact [7]

Answer:

480.32 foot per second

Explanation:

there is a conversion thing ona famous search engine use that and I just multiple by 4

8 0
3 years ago
A charged box (m=445 g, ????=+2.50 μC) is placed on a frictionless incline plane. Another charged box (????=+75.0 μC) is fixed i
victus00 [196]

The concept required to perform this exercise is given by the coulomb law.

The force expressed according to this law is given by

F= \frac{kqQ}{r^2}

Where,

k = 8.99 * 10^9 N m^2 / C^2.

q = charges of the objects

r= distance/radius

Our values are previously given, so

q= 2.5*10^{-6}C\\Q= 75*10^{-6}C\\r=0.59

Replacing,

F=\frac{kqQ}{r^2}

F= \frac{(8.99 x 10^9)(2.5*10^{-6})(75*10^{-6})}{0.59^2}

F= 4.8423N

The force acting on the block are given by,

F-mgsin\theta = ma

a = \frac{F-mgsin\theta}{m}

a = \frac{4.8423-(0.445)(9.8)sin(35)}{0.445}a = 10.31m/s^2

Therefore the box is accelerated upward.

3 0
3 years ago
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